Two particles have equal masses of 5.0 g each and opposite charges of `+4.0 xx10^(-5) C.` They are released from rest with a separation of 1.0 m between them. Find the speeds of the particles when the separation is reducced to 50 cm.

Explanation

`q_1=q_2=4xx10^-5,S=1m, `
` m=5g=0.005kg, `
` f=k=14.4N `
` So accln a=2880m//s^2 `
` Now, u=0, s=50cm=0.5m, `
` a=2880m//s^2, v^2=? `
` v^2=u^2+2as `
` v62=2xx2880xx0.5 `
` v=sqrt(2280)=53.66m//s `
` =54m//s for each particle `